Question: $f$ is the Maclaurin series $\sum_{n=0}^{\infty}\frac{{{x}^{2n}}}{n!}$. $f\left(\sqrt{ \ln(3)}\right)=$
Solution: We can use the Maclaurin series for ${{e}^{x}}$ as a guide. Recall that ${{e}^{x}}=\sum_{n=0}^{\infty}\frac{{{x}^{n}}}{n!}~$. Hence, $\sum_{n=0}^{\infty}\frac{{{x}^{2n}}}{n!}~=\sum_{n=0}^{\infty}\frac{{{({{x}^{2}})}^{n}}}{n!}={{e}^{{{x}^{2}}}}$. Now substitute the given value of $x\,$. $f(\sqrt{\ln(3)})= {{e}^{{{\left(\sqrt{\ln(3)}\right)}^{2}}}} = {{e}^{\ln 3}} = 3\,$